DO NOW:
Do Now (2/18/14):
 Pass in your hw, lab, and Do Now!!!
 A concave mirror with a radius of curvature
of 1.0 m is used to collect light from a distant
star. The distance between the mirror and
the image of the star is most nearly
a) 0.25 m
b) 0.50 m
c) 0.75 m
d) 1.0 m
e) 2.0 m
Objectives
 Describe the causes of interference patterns due to
diffraction and thin films.
 Use the diffraction equation.
 Describe the effects of polarization of light.
 Determine the wavelength of a light source using a
diffraction pattern.
Brainstorm:
 Discuss with your elbow partner (2 min):
 Do you think light is a wave or a
particle? Why or why not?
Intensity
 Is the energy it carries per unit of time.
 Proportional to the square of the amplitude of the
wave.
 Color of light is related to it’s λ or f, not intensity.
 What characteristic of sound is most similar to light
intensity?
Polarization
 Polarizing filters
 Only possible if light travels as a wave.
 I = I0cos2(θ) (Malus’s Law)
Huygen’s Principle
 In the 17th Century,
Christian Huygens
proposed a theory
stating: every point on a
wave front can be
considered as a source
of tiny wavelets that
spread out in the
forward direction at the
speed of the wave.
Huygen’s Principle
 Every point of a wave front may be
considered the source of secondary wavelets
that spread out in all directions with a speed
equal to the speed of propagation of the
waves.
Huygens’ Principle
 Other scientists immediately
understood that Huygens’ principle
predicted that all waves should
then spread into the shadow
behind an obstacle.
 We call this bending of waves
diffraction.
 If light is a wave, it should undergo
the process of diffraction, and
should also be able to undergo the
process of interference.
Light: Particle or Wave?
 A light source illuminating a single slit
http://images.google.com/imgres?imgurl=http://www.peace-files.com/QF-L-11/07_Double-Slit-03.jpg&imgrefurl=http://www.peace-files.com/QF-L-11/01_QF-Doubleslit.html&h=267&w=400&sz=26&hl=en&start=68&um=1&tbnid=K6ikhnty_r8L4M:&tbnh=83&tbnw=124&prev=/images%3Fq%3Dyoung%2527s%2Bdouble%2Bslit%26start%3D54%26ndsp%3D18%26um%3D1%26hl%3Den%26rls%3Dcom.microsoft:*%26sa%3DN
Light: Particle or Wave?
 Particle theory prediction for the pattern produced
by two slits, side by side:
http://images.google.com/imgres?imgurl=http://www.peace-files.com/QF-L-11/07_Double-Slit-03.jpg&imgrefurl=http://www.peace-files.com/QF-L-11/01_QF-Doubleslit.html&h=267&w=400&sz=26&hl=en&start=68&um=1&tbnid=K6ikhnty_r8L4M:&tbnh=83&tbnw=124&prev=/images%3Fq%3Dyoung%2527s%2Bdouble%2Bslit%26start%3D54%26ndsp%3D18%26um%3D1%26hl%3Den%26rls%3Dcom.microsoft:*%26sa%3DN
Light: Particle or Wave?
 Wave theory prediction for the pattern produced by two slits,
side by side:
 Note: to easily observe light-wave interference, light from the
two sources would have to be coherent and monochromatic.
http://images.google.com/imgres?imgurl=http://www.peace-files.com/QF-L-11/07_Double-Slit-03.jpg&imgrefurl=http://www.peace-files.com/QF-L-11/01_QF-Doubleslit.html&h=267&w=400&sz=26&hl=en&start=68&um=1&tbnid=K6ikhnty_r8L4M:&tbnh=83&tbnw=124&prev=/images%3Fq%3Dyoung%2527s%2Bdouble%2Bslit%26start%3D54%26ndsp%3D18%26um%3D1%26hl%3Den%26rls%3Dcom.microsoft:*%26sa%3DN
Young’s Double Slit Experiment
 Young saw an
interference pattern.
 To explain this
alternating pattern of
bright and dark fringes,
he understood that at
any point on the viewing
screen (other than at the
center), the two rays of
light would have to
travel different
distances to arrive at
the screen.
x
d
ϴ
L
Path Difference
Young’s Double Slit Experiment
 If this path length
difference is an integer
multiple of the light’s
wavelength, a bright
fringe is seen.
d
ϴ
d sin   m
 A dark fringe is seen
anytime the path length
difference is ½ of the λ.
d sin   m  1 2 
dsinϴ
Fringes
 Diffraction causes
fringes:
Example Problem 24-1
 520nm light falls on a pair of narrow slits separated by
0.2mm. How far apart are the fringes near the center of
the pattern on a screen 2.5m away?
x = 0.064m or 6.4mm
Small Angle Approximation
 How could we approximate Young’s formula for
small angles??
Small Angle Approximation
 Note: For small angles (x << L), sinϴ ≈ tanϴ, so you
can approximate the distance between the fringes
as…
Diffraction Gratings
 A special device called a diffraction grating consists of
many equally-spaced parallel lines scratched into a glass
plate. The spaces between the scratches act as a source of
light, and interference is observed.
 The interference equation predicts the location of
maxima in the interference pattern. The maxima are
much thinner & more defined than the pattern created
by a double slit.
 The only difference about these problems (compared to
old diffraction problems) will be that you won’t be told
the value of d. Instead, you’ll be told the number of lines
per distance, which is really 1/d.
Review:
Huygens’ principle
 Wave theory of light – every point on a wave front
can be considered as a source of tiny wavelets that
spread out in the forward direction at the speed of
the wave itself.
 predicts waves bending around openings
 http://www.launc.tased.edu.au/online/sciences/phy
sics/diffrac.html
Diffraction
 Young’s Double slit experiment – wave nature of
light
 dsinθ = m - constructive interference
 xm= (m L)/d
http://www.surendranath.org/applets/optics/slits/doubleslit/dblsltapplet.html
http://micro.magnet.fsu.edu/primer/java/interference/doubleslit/
 Plane sound waves of wavelength 0.12 m are incident on
two narrow slits in a box with nonreflecting walls, as
shown above. At a distance of 5.0 m from the center of
the slits, a first order maximum occurs at point P, which
is 3.0 m from the central maximum. The distance
between the slits is most nearly
a) 0.07 m
b) 0.09 m
c) 0.16 m
d) 0.20 m
e) 0.24 m
Single Slit Diffraction
 It turns out that an
Numerical Approximation
interference pattern is still
observed even when there’s
only one point as a source
of light.
 According to Huygens’
principle, each portion of
the slit acts as a source of
waves. Therefore, light
from one portion of the slit
can interfere with light
from a different portion.
Single Slit Diffraction
 The only thing that
changes about problems
with single-slit diffraction
is that the interference
equation now predicts the
location of minima
(where the diffraction
pattern has minimum
intensity), for integral mvalues.
d sin   m
Different Diffraction Patterns
http://www.tau.ac.il/~phchlab/experiments/hydrogen/diffraction_gratings.ht
m
Diffraction Patterns from Edges
 Diffraction patterns can arise anytime light bends
when passing around edges of an obstacle.
 Shadows of objects therefore contain diffraction
patterns, with a bright spot at their centers.
 You should also be aware that light wave diffraction
isn’t observed as much in the macroscopic (big)
world because diffraction effects are more
pronounced when the size of the opening through
which the wave passes is close to the size of the
wavelength of the wave.
Do Now (2/19/14):
 If the distance between two slits is 0.050 mm and
the distance to a screen is 2.5 m, find the spacing
between the first and second order bright fringes for
light of wavelength 600 nm.
 What color is this light?
Single Slit Diffraction
 Diffraction
of light by a
slit of
narrow
width a
Example:
 Light of wavelength 580 nm is incident on a slit of
width 0.3 mm. The observing screen is placed 2 m
from the slit. Find the positions of the first dark fringes
and the width of the central bright fringe.
3
 1.9 x10
3
 3.9 x10 m
Brewster’s Law:
Polarizing Angle
 The angle of incidence that satisfies Brewster’s Law.
n p  tan p
Thin-Film Interference
Thin Film Interference
 When light encounters the
boundary between two
substances with different
indices of refraction , some of
the light will be reflected and
some will be transmitted.
 Interference happens.
 Constructive interference
happens when the 2nd wave
exits the whole mess in phase
with the 1st wave.
 Depth of film = 1/2λ
Air (n=1.00)
Oil (n=1.28)
Water (n=1.33)
Thin Film Interference
 Destructive interference
happens when the 2nd wave
exits the whole mess out of
phase with the 1st wave.
 Depth of film = 1/4λ
Air (n=1.00)
Oil (n=1.28)
Water (n=1.33)
Thin-Film Interference
 When light is
reflected upon trying
to enter a substance
with a higher n-value,
it is also shifted by
180°, which is equal
to a path difference of
1/2λ.
Air (n=1.00)
Glass (n=1.56)
Glass (n=1.56)
Air (n=1.00)
Thin-Film Interference
 Constructive interference will
Air (n=1.00)
Soap (n=1.28)
Air (n=1.00)
happen when the second wave
undergoes a total phase change of λ.
 But since during reflection it
undergoes a phase change of 1/2λ
when it reflects from the top soap
layer, it only needs to undergo a
phase change of 1/2λ as it travels the
thickness of the film (twice).
Thin Film Interference
 Constructive:
1

2nt   m   , m  (0,1,2,...)
2

 Destructive:
2nt  m
Thin Film Interference
 The wavelength of light, λn, in a medium with index
of refraction n is:

n 
n
 Where λ is the wavelength of light in free space
Example:
 Calculate the minimum thickness of a soap-bubble
film (n=1.33) that will result in constructive
interference in the reflected light if the film is
illuminated by light with a wavelength in free space
of 602 nm.
113 nm
AP Practice!
 Try to finish the first two problems before the end of
today.
Lab: How wide is a human hair?
 A human hair can act just like a double slit. The light
going around both edges will bend/diffract and
create an interference pattern.
 This lab will include a writeup
Electromagnetic Waves and
Optics
AP PHYSICS
UNIT 11
GIANCOLI
CH.22 - 24
Electromagnetic Waves
 We already know that a
changing B-field or flux will
produce an electric field (i.e.
causes the movement of
charges or current)
 Conversely, James Maxwell
came up with the idea that a
changing electric field can
produce a magnetic field.
Electromagnetic Waves
 Accelerating charge gives rise to E&M
waves that can even travel through a
vacuum.
 The oscillating electric and magnetic
fields are perpendicular to one another.
 E&M waves move through a vacuum at
c = 3.00x108 m/s.
Electromagnetic Waves
 The wave-speed equation still applies:
c  f
Speed of Light (c)
 Ole Roemer first determined that
the speed of light was finite.
 He found that the period of Io,
one of Jupiter’s moons, varied
slightly depending on the relative
motion of Earth and Jupiter.
 If the Earth was moving away
from Jupiter during Io’s orbit the
light would have to travel a longer
distance, increasing I0’s apparent
orbital period.
Electromagnetic Spectrum
Inverse Square Relationship
Ch.22 Homework
Read sections 22.1 – 22.2 (no math in
either), 22.3-22.4 and 22.7
Questions: 1, 3, 5, 10 & 13
Problems: 5-6, 9 & 16
Due: Tomorrow
Ch.22 Homework Answers
 5. 1.88E10 Hz
 6. 1.008E-10 m
 9. 8.33min
 16. radio hears 0.14s soon
Physics of Sight
 We see an object in one of two ways:
 The object is a source of light (sun, fire, light bulb filament)
 The object reflects light
 Reflected Light rays scatter from each point on an object.
 Our brains construct the image of an object assuming that the light
entering our eyes travels in straight lines.
Law of Reflection
 The angle of reflection is equal to the angle of
incidence.
1   2
ϴ1
ϴ2
ϴ1
ϴ2
Plane Mirrors
 The image formed by a plane mirror is a virtual
image (cannot be projected onto a screen)
Spherical Mirrors
 Concave mirrors reflect
incoming parallel light
rays so that they pass
through a common focal
point
 Convex mirrors reflect
incoming parallel light
rays so that they appear
like they are coming
from a focal point
behind the mirror.
Spherical Mirrors
 C: Center of curvature
 F: focal point
 r: radius of curvature
C
F
 f: focal length
r
f 
2
F
f
C
Spherical Aberration
 Technically speaking
spherical mirrors do not
focus the rays perfectly.
And the more spherical a
mirror is, the more the
image will appear
blurred. This defect is
called spherical
aberration. For very
sensitive applications
parabolic mirrors are
used.
http://wisp.physics.wisc.edu/astro104/lecture7/F06_13.jpg
Concentrating Solar Power (CSP) Plants
 The sun’s rays are focused
on pipe filled a fluid to
collect the energy in order
to generate electricity.
http://en.wikipedia.org/wiki/Solar_thermal_energy#HighTemperature_Collectors:_Concentrated_solar_power_.28CSP.29_plants
http://images.google.com/imgres?imgurl=http://www.flabeg.com/images/g_03_solar_mirrors.jpg&imgrefurl=http://www.flabeg.com/en/03_solar_mirrors.html&
h=385&w=385&sz=25&hl=en&start=120&um=1&tbnid=2JjpA8c5Og3NrM:&tbnh=123&tbnw=123&prev=/images%3Fq%3Dparabolic%2Bmirror%26start%3D1
08%26ndsp%3D18%26um%3D1%26hl%3Den%26rls%3Dcom.microsoft:*%26sa%3DN
Drawing Ray Diagram Rules
 Rule #1: Draw a ray going out from the object
parallel to the principal axis that reflects back
through the focal point.
Drawing Ray Diagram Rules
 Rule #2: Draw a ray that goes through the focal point
(or in a direction like it came from the focal point)
then reflects back parallel to the axis.
Drawing Ray Diagram Rules
 Rule #3: Draw a ray from the object through the
center of curvature. This ray will strike the mirror at
a right angle and will reflect back on itself.
r >object distance > f
focal point is +
Image is…
Real, larger & inverted
si (+)
so (+)
object distance < f
so (+)
Image is…
Virtual, larger & upright
si (-)
Ray Diagram for Convex Mirror
Image is always…
Virtual, smaller & upright
so (+)
si (-)
focal point is (-)
Uses: rear view mirror, convenience store mirror
The Mirror Equation
1 1 1
 
f so si
si hi
M  
so ho
 h0 is always (+)
 hi is (+) if upright, (-) if inverted with respect to the
object
 (+)si  image in front of mirror
(-) si  image behind mirror
 f (+) for concave mirrors, (-) for convex mirrors
Example Problem 23-4
 A 10cm-tall object is placed 12cm in front of a convex mirror that
has a radius of curvature of 35cm. Completely describe the
reflected image. (What is its location? Its height? Is it real or
virtual? Upright or inverted?)
si = -7.1cm, hi = 5.92cm, virtual and upright
Example Problems 23-4 and 23-5
 A concave mirror with a radius of curvature of 14cm is used to
focus the Sun’s rays. Where are the rays focused, relative to the
surface of the mirror?
7cm from the surface of the mirror at the focal point
 You are standing 3.0m from a convex security mirror in a store.
You estimate the height of your image to be half of your actual
height. Estimate the focal point of the mirror.
f = -3m
Refraction
 All lenses redirect light rays by the process of
refraction:
n1
n2
ϴ2
c
n
v
ϴ1
n1 sin 1  n 2 sin  2
Lenses
 Lenses can be grouped into two main categories:
converging and diverging lenses.
Ray Diagrams for Lenses…
Example Problem
 Use a ray diagram to show the image height and
position for the given object.
F
F
Thin Lens Calculations
 The mirror equation still applies, but now it’s called




the thin lens equation. (The magnification equation
is still the same, too.)
Sign conventions for using the equations are
somewhat different now…
s0 is (+) if it is on the same side of the lens as the
incoming light… which is most of the time.
If si is (+) then the image is on the opposite side of
the incoming light, (-) if on the same side.
f is (+) for converging lenses, (-) for diverging lenses.
Multiple Lenses
 If two thin lenses in a row are used to form an image,
first find the image of the first lens alone. Then the
light approaches a 2nd lens as if it had come from the
image. This means that the image formed by lens #1
becomes the object for lens #2.
F
F
F
F
Multiple Lenses
 The magnification for multiple lenses is just the
product of the individual magnifications of lens #1 &
#2.
 (Example) Two converging lenses, each of focal
length 20cm, are place 50cm apart from one
another, and an object is place 10cm to the left of the
first lens. Where is the final image formed, and what
is the magnification of the entire system?